字符串处理 Codeforces Round #285 (Div. 2) B. Misha and Changing Handles

题目传送门

/*
    题意：给出一系列名字变化，问最后初始的名字变成了什么
    字符串处理：每一次输入到之前的找相印的名字，若没有，则是初始的，pos[m] 数组记录初始位置
                在每一次更新时都把初始pos加上去，那么就保证更新了初始的名字，这也是唯一要思考的地方了:)
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <ctime>
using namespace std;

const int MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
string old[MAXN];
string change[MAXN];
string ans[MAXN];
int pos[MAXN];

int main(void)        //Codeforces Round #285 (Div. 2) B. Misha and Changing Handles
{
    freopen ("B.in", "r", stdin);

    int n;
    while (scanf ("%d", &n) == 1)
    {
        memset (pos, 0, sizeof (pos));
        string tmp;    int m = 0;
        for (int i=1; i<=n; ++i)
        {
            cin >> tmp; cin >> change[i];
            bool ok = false;
            for (int j=1; j<i; ++j)
            {
                if (tmp == change[j])
                {
                    ok = true;
                    ans[pos[j]] = change[i];
                    pos[i] = pos[j];    break;
                }
            }
            if (!ok)
            {
                old[++m] = tmp; ans[m] = change[i];    pos[i] = m;
            }
        }

        printf ("%d
", m);
        for (int i=1; i<=m; ++i)
        {
            cout << old[i] << " " << ans[i] << endl;
        }
    }

    // cout << "time: " << (double) clock () / CLOCKS_PER_SEC * 1000 << " ms
";

    return 0;
}